the enol form of `hept-5-en-2-`one contains

Options

A. 17sigma-bonds and 2π-bonds
B. 19sigma-bonds and 2π-bonds
C. 15sigma-bonds and 1π-bond
D. 21sigma-bonds and 2π-bonds

Answer

To determine the enol form of hept-5-en-2-one, we will follow these steps:

### Step 1: Identify the Structure of Hept-5-en-2-one
Hept-5-en-2-one has 7 carbon atoms (hept) and a double bond starting at the 5th carbon (5-en) with a ketone functional group at the 2nd carbon (2-one). The structure can be drawn as follows:

```
CH3-CH2-C(=O)-CH=CH-CH2-CH3
```

### Step 2: Draw the Enol Form
To convert the ketone to its enol form, we need to perform a tautomerization reaction. This involves the migration of a hydrogen atom and the conversion of the carbonyl group (C=O) into an alcohol (C-OH) while forming a double bond between the adjacent carbon atoms.

1. Move one hydrogen from the 3rd carbon (C3) to the oxygen of the carbonyl group (C=O).
2. The double bond between C4 and C5 will shift to create a double bond between C2 and C3.

The enol form can be represented as:

```
CH3-CH=C(OH)-CH=CH-CH2-CH3
```

### Step 3: Count Sigma and Pi Bonds
Now, we will count the sigma (σ) and pi (π) bonds in the enol form.

- **Sigma Bonds**: Each single bond counts as one sigma bond, and each double bond counts as one sigma bond plus one pi bond.
- **Pi Bonds**: Each double bond counts as one pi bond.

In the enol structure:
- The sigma bonds can be counted as follows:
- C1 to C2: 1
- C2 to C3: 1
- C3 to C4: 1
- C4 to C5: 1
- C5 to C6: 1
- C6 to C7: 1
- C1 to H: 3 (3 H atoms on C1)
- C2 to H: 1 (1 H atom on C2)
- C3 to H: 1 (1 H atom on C3)
- C4 to H: 1 (1 H atom on C4)
- C5 to H: 2 (2 H atoms on C5)
- C6 to H: 2 (2 H atoms on C6)

Total sigma bonds = 19

- **Pi Bonds**: There are 2 double bonds in the enol form.
Total pi bonds = 2

### Final Answer
The enol form of hept-5-en-2-one contains **19 sigma bonds and 2 pi bonds**.

---